Problem: Solve for $x$ : $2x^2 + 4x - 160 = 0$
Dividing both sides by $2$ gives: $ x^2 + {2}x {-80} = 0 $ The coefficient on the $x$ term is $2$ and the constant term is $-80$ , so we need to find two numbers that add up to $2$ and multiply to $-80$ The two numbers $10$ and $-8$ satisfy both conditions: $ {10} + {-8} = {2} $ $ {10} \times {-8} = {-80} $ $(x + {10}) (x {-8}) = 0$ Since the following equation is true we know that one or both quantities must equal zero. $(x + 10) (x -8) = 0$ $x + 10 = 0$ or $x - 8 = 0$ Thus, $x = -10$ and $x = 8$ are the solutions.